Chapter 5, Lect. 9Additional Applications of Newton’s Laws

Today: Fractional forces, drag forces

04/20/23 Phys 201, Spring 2011

• When and where– Thurs Feb. 17th 5:45-7:00 pm– Rooms: See course webpage. Be sure report to your TA’s room – Your TA will give a review during the discussion session next week.

• Format– Closed book, 20 multiple-choices questions (consult with practice exam)– 1page 8x11 formula sheet allowed, must be self prepared, no photo

copying/download-printing of solutions, lecture slides, etc.– Bring a calculator (but no computer). Only basic calculation functionality can be used. Bring a 2B pencil for Scantron.– Fill in your ID and section # !

• Special requests: – One alternative exam all set: 3:30pm – 4:45pm, Thurs Feb.17, room 5280 Chamberlin (!).

About Midterm Exam 1

Frictional Force Friction

Opposes motion between systems in contactParallel to the contact surfaceDepends on the force holding the surfaces together the normal force N and a constant: coefficient of friction μ

Static frictionFrictional force without relative motion

fs is less than or equal to μs N Kinetic friction

Frictional force on an object in motionCan be less than static friction

fk = μk N

04/20/23 Phys 201, Spring 2011

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fs ≤μsN

fk =μkN

Application: ABS(Anti-lock Brake System)

Anti-lock brakes work by making sure the wheels roll without slipping. This maximizes the frictional force slowing the car since μS > μK .

04/20/23 Phys 201, Spring 2011

Surface friction…

maaFF

ffFmgg

NN

ii

j j

Force of friction acts to oppose relative motion:Parallel to surface.Perpendicular to normal forceIt is determined (proportional to) normal force.

friction force04/20/23 Phys 201, Spring 2011

v0 =8 m/sμk= 0.2

Initial speed v0: Find stopping distance

Normal force is balanced by gravity because there is no vertical motion: N = Mg, if M is the mass of the object

Kinetic frictional force that decelerates the block is, f = μk N = μk Mg

Therefore, deceleration (direction opposite of v0), a = -f/M = -μk g

Given deceleration, use kinematics equation to obtain the answer.

Answer: 16.3 m

-- independent of the mass

-- the smaller μk is, the larger the stopping distance will be.

Example 1N

Mg

μκ N

04/20/23 Phys 201, Spring 2011

Example 2T=50 N

M = 5 kg

μk = 0.2

Find acceleration of the block M

θ=500

04/20/23 Phys 201, Spring 2011

Draw free body diagramResolve T in x and y components: Tx=T sinθ, Ty=T cosθ

Solve for y-component of force: No vertical motion: N + Ty = Mg Solve for x-component of force: Fx = Tx – fs , with fs = μk N

Then use ax=Fx/m

Answer: acceleration of block is 6.0 m/s2 in +x direction

Example 3: Consider on an inclined plane, with an angle θ, w.r.t the horizontal.

In this case, the force provided by friction will depend on the angle θ of the plane:

because of the normal force

θ

04/20/23 Phys 201, Spring 2011

Inclined Plane with Friction:

θ

ii

jj

mgN

μK Nma

θ

Free-body diagram for block:

04/20/23 Phys 201, Spring 2011

Consider i and j components of FNET = ma

θ

ii

jj

mg

N

θ

μK N

ma

mg sin θmg cos θ

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i coμponent : μ g sinθ -μKN=μaj coμponent: N=μg cosθ

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μg sinθ−μKμg cosθ =μa

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ag=sinθ−μKcosθ

04/20/23 Phys 201, Spring 2011

If a = 0 Static friction!

As long as μs > tan θ

Is the frictional force always oppositeto the moving direction?

Friction keeps the car wheels from spinning in placeYou want the tires to roll, clock-wise to your view:Friction opposes to itThe contact point is at rest - although the car is in motion

» What matters is the coefficient of static friction!

weight Maximum Static friction ( > Fcar on road for car not to spin in place!)

Fcar on road

Froad on car

Consider Newton’s 3rd law:

Froad on car is the actual force ON the car.

Static Friction μsN is its maximum value

04/20/23 Phys 201, Spring 2011

A kid on a toboggan

04/20/23 Phys 201, Spring 2011

Naively, Child: ftc = mc aToboggan (slippery ground): F - ftc = mt a

Thus: F = mc a + mt a a = F/(mc+mt)However, the static frictional force cann’t help forever! ftc

max = μs mc g Then the maximum acceleration: mc amax = ftc

max Thus: a = F/(mc+mt) ≤ μs g otherwise … …

Drag Forces

• Objects moving through a fluid such as air or water experience a drag force that opposes the motion of the object. The magnitude of the drag force increases as the speed increases (unlike the kinetic friction force!). Empirically it is typically found that

04/20/23 Phys 201, Spring 2011

where the coefficient b is a constant, depends mainly on crossing area.n is typically 1 – 2.

“Terminal speed” of falling object

• The terminal speed vT is the speed at which the drag force bvn exactly balances the force of gravity mg.

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bvTn =μg

vT = μgb

⎛

⎝ ⎜

⎞

⎠ ⎟1/ n

04/20/23 Phys 201, Spring 2011

Question – sky diverA sky diver jumps out of an airplane at 5000m altitude. She reaches terminal speed (due to the drag of the air) after about six seconds.

If a box of steel parts that has the same weight as the diver is dropped simultaneously, the box will fall:

faster than the diver

slower than the diver

the same as the diver

The force of gravity is proportional to the mass of an object, and the drag coefficient is proportional to the cross-section of the object. The box of steel is much more dense than a human body (whose density is about that of water), so the crossing area (volume too) of the box is much smaller compared to that for the person. In other words, the terminal velocity vT=(mg/b)1/n, which is larger for the box of steel than it is for the person because of b.

04/20/23 Phys 201, Spring 2011